Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using cd = complex<double>;
const int N = 3e3 + 5;
const long long INF = 1e18;
const int mod = 998244353;
const double Pi = acos(-1);
const int p = 510200;
void Fastio() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
int n, m, q, tim = 0;
int tin[100005], tout[100005];
int Par[100005][18];
vector<int> Adj[100005], Up[100005][18], T[100005];
bool Used[100005];
vector<int> Merge(vector<int> A, vector<int> B) {
  vector<int> Temp = A;
  int p = 0;
  for (auto x : A) {
    Used[x] = true;
  }
  for (auto x : B) {
    if (Used[x]) {
      continue;
    }
    Temp.emplace_back(x);
  }
  sort(Temp.begin(), Temp.end());
  while (Temp.size() > 10) {
    Temp.pop_back();
  }
  for (auto x : A) {
    Used[x] = false;
  }
  return Temp;
}
void DFS(int node, int p = 0) {
  tim++;
  tin[node] = tim;
  Par[node][0] = p;
  Up[node][0] = Merge(T[node], T[p]);
  for (int i = 1; i <= 17; i++) {
    Par[node][i] = Par[Par[node][i - 1]][i - 1];
    Up[node][i] = Merge(Up[node][i - 1], Up[Par[node][i - 1]][i - 1]);
  }
  for (auto x : Adj[node]) {
    if (x != p) {
      DFS(x, node);
    }
  }
  tout[node] = tim;
}
bool Isparent(int u, int v) { return tin[u] <= tin[v] and tout[u] >= tout[v]; }
int LCA(int u, int v) {
  if (Isparent(u, v)) {
    return u;
  }
  if (Isparent(v, u)) {
    return v;
  }
  for (int i = 17; i >= 0; i--) {
    if (!Isparent(Par[u][i], v)) {
      u = Par[u][i];
    }
  }
  return Par[u][0];
}
signed main() {
  Fastio();
  cin >> n >> m >> q;
  for (int i = 1; i <= n - 1; i++) {
    int u, v;
    cin >> u >> v;
    Adj[u].emplace_back(v);
    Adj[v].emplace_back(u);
  }
  tin[0] = 0;
  tout[0] = 1e9;
  for (int i = 1; i <= m; i++) {
    int temp;
    cin >> temp;
    if (T[temp].size() < 10) {
      T[temp].emplace_back(i);
    }
  }
  DFS(1);
  while (q--) {
    int u, v, a;
    cin >> u >> v >> a;
    int temp = LCA(u, v);
    vector<int> ans;
    for (int i = 17; i >= 0; i--) {
      if (!Isparent(Par[u][i], temp)) {
        ans = Merge(ans, Up[u][i]);
        u = Par[u][i];
      }
    }
    for (int i = 17; i >= 0; i--) {
      if (!Isparent(Par[v][i], temp)) {
        ans = Merge(ans, Up[v][i]);
        v = Par[v][i];
      }
    }
    ans = Merge(ans, T[v]);
    ans = Merge(ans, T[u]);
    ans = Merge(ans, T[temp]);
    cout << min(int(ans.size()), a) << ' ';
    for (int i = 0; i < min((int)ans.size(), a); i++) {
      cout << ans[i] << ' ';
    }
    cout << '\n';
  }
}